Given below are the number of successes and sample size for a simple random sample from a population. xequals6​, nequals50​, 90​% level a. Determine the sample proportion. b. Decide whether using the​ one-proportion z-interval procedure is appropriate. c. If​ appropriate, use the​ one-proportion z-interval procedure to find the confidence interval at the specified confidence level. d. If​ appropriate, find the margin of error for the estimate of p and express the confidence interval in terms of the sample proportion and the margin of error.

Accepted Solution

Answer:a. Sample proportion ^p= 0.12b. It is appropiate.c. [0.0447;0.1953]d. [^p ± d]Step-by-step explanation:Hello!Given the information I'll assume that the variable of study has a binomial distribution:X~Bi(n;ρ)The sample data:n= 50"Success" x= 6Sample proportion ^p= x/n = 6/50 = 0.12Now, your study variable has a binomial distribution, but remember that the Central Limit Theorem states that given a big enough sample size (usually n≥ 30) you can approximate the sample proportion distribution to normal.Since the sample is 50 you can apply the approximation, your sample proportion will have the following distribution:^p≈ N( p; [p(1 - p)]/n)With E(^p)= p and V(^p)= [p(1 - p)]/n.This allows you to estimate the population proportion per Confidence Interval using the Z-distribution:[^p±[tex]Z_{1-\alpha /2}[/tex]*√(^p(1 - ^p)/n)]Since you are estimating the value of p, you'll use the estimated standard deviation (i.e. with the sample proportion instead of the population proportion)to calculate the interval.At level 90% the interval is:[0.12±1.64*√([0.12(1 - 0.12)]/50)][0.0447;0.1953]The margin of error (d) of an interval is half its amplitude (a)if a= Upper bond - Low bondthen d= (Upper bond - Low bond)/2d= (0.1953-0.0447)/2d= 0.0753And since the interval structure is "estimator" -/+ "margin of error" you can write it as:[^p ± d]I hope you have a SUPER day!